2 replies to this topic

[muamuaha125]

1.Cho x;y;z>0 thoã mãn: $x+y+z+\sqrt{xyz}=4$. Tính: P=$\sqrt{x(4-y)(4-z)}+\sqrt{y(4-x)(4-z)}+\sqrt{z(4-y)(4-x)}-\sqrt{xyz}$

2. Cho a;b thoả mãn b>$\frac{a^2}{4}$. CMR: $\sqrt[3]{\frac{ab+\sqrt{a^2b^2-\frac{4}{27}(a^2-b)^3}}{2}}+\sqrt[3]{\frac{ab-\sqrt{a^2b^2-\frac{4}{27}(a^2-b)^3}}{2}}=a$

3. Cho $\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}=a+b+c$. Tính P=$\frac{a^2+b^2}{(a+c)(b+c)}+\frac{b^2+c^2}{(b+a)(c+a)}+\frac{c^2+a^2}{(c+b)(a+b)}$

4. Cho x=$\sqrt{2}+\sqrt[3]{3}$. Tính P=$x^{2013}-6x^{2011}-6x^{2010}+12x^{2009}-36x^{2008}+x^{2007}+2013$

[Zaraki]

4. Cho x=$\sqrt{2}+\sqrt[3]{3}$. Tính P=$x^{2013}-6x^{2011}-6x^{2010}+12x^{2009}-36x^{2008}+x^{2007}+2013$

Lời giải. Ta có $x- \sqrt 2= \sqrt[3]{3} \Rightarrow x^3-3\sqrt 2 x^2+6x-2 \sqrt 2= 3 \Rightarrow x^3+6x-2=(3x^2+2) \sqrt{2}$

$\Rightarrow (x^3+6x-2)^2=2(3x^2+2)^2 \Rightarrow x^6-6x^4-6x^3+12x^2-36x+1=0$.

Ta có $P=x^{2007} \left( x^6-6x^4-6x^3+12x^2-16x+1 \right)+2013=2013$.

[Rikikudo1102]

ta xét $\sqrt{x(4-y)(4-z)}=\sqrt{x(16-4y-4z+yz)}$ (*)

 

ta có: $x+y+z+\sqrt{xyz}=4$ ---> $16-4y-4z=4x+4\sqrt{xyz}$

 

thay vào (*) ta có $\sqrt{x(16-4y-4z+yz)}=\sqrt{x(4x+4\sqrt{xyz})}$

 

=$\sqrt{4x^2+4x\sqrt{xyz}+xyz}=2x+\sqrt{xyz}$

 

tt: $\sqrt{y(4-x)(4-y)}=2y+\sqrt{xyz}$

 

$\sqrt{z(4-x)(4-y)}=2z+\sqrt{xyz}$

 

---> P= 8

Edited by Rikikudo1102, 18-11-2013 - 20:15.